标准正态分布k阶原点矩公式

标准正态分布k阶原点矩公式

今天在看 χ 2 \chi^2 χ2分布,计算其方差时,遇到了标准正态分布的四阶原点矩。书上直接写 E ( X i 4 ) = 3 E(X_i^4)=3 E(Xi4)=3,很好奇。设 X i ∼ N ( 0 , 1 ) X_i\sim\mathcal{N}(0,1) XiN(0,1)想根据定义计算:
E ( X i 4 ) = ∫ − ∞ + ∞ x 4 1 2 π e − x 2 2 d x E(X_i^4)=\int_{-\infty}^{+\infty}x^4\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx E(Xi4)=+x42π 1e2x2dx
计算起来复杂度有点高,K阶就更不敢想。
想着估计结果是 k k k的递推关系式。所有直接计算:
E ( X k ) = ∫ − ∞ + ∞ x k 1 2 π e − x 2 2 d x = 1 k + 1 x k + 1 1 2 π e − x 2 2 ∣ − ∞ + ∞ + 1 k + 1 ∫ − ∞ + ∞ x k + 2 1 2 π e − x 2 2 d x = 0 + 1 k + 1 E ( X k + 2 ) \begin{aligned} E(X^k) &= \int_{-\infty}^{+\infty}x^k\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx \\ & = \frac{1}{k+1}x^{k+1}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} |_{-\infty}^{+\infty} + \frac{1}{k+1}\int_{-\infty}^{+\infty}x^{k+2}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx \\ & = 0 + \frac{1}{k+1}E(X^{k+2}) \end{aligned} E(Xk)=+xk2π 1e2x2dx=k+11xk+12π 1e2x2++k+11+xk+22π 1e2x2dx=0+k+11E(Xk+2)
从上式看,结果已经很明显了,其递推关系为:
E ( X k ) = ( k − 1 ) E ( X k − 2 ) , k = 2 , 3 , 4 ⋯ E(X^k) = (k-1)E(X^{k-2}),k=2,3,4\cdots E(Xk)=(k1)E(Xk2)k=2,3,4
其中:
E ( X 0 ) = ∫ − ∞ + ∞ x 0 1 2 π e − x 2 2 d x = 1 E(X^0)=\int_{-\infty}^{+\infty}x^0\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=1 E(X0)=+x02π 1e2x2dx=1
E ( X 1 ) = 0 E(X^1)=0 E(X1)=0
所以,当 k = 2 i k=2i k=2i为偶数时:
E ( X k ) = ( k − 1 ) ( k − 3 ) ⋯ 3 × 1 × E ( X 0 ) = ∏ i = 1 k / 2 ( 2 i − 1 ) \begin{aligned} E(X^k) &= (k-1)(k-3)\cdots3\times1\times E(X^0) \\ &=\prod_{i=1}^{k/2}(2i-1) \end{aligned} E(Xk)=(k1)(k3)3×1×E(X0)=i=1k/2(2i1)
k = 2 i − 1 k=2i-1 k=2i1为奇数时:
E ( X k ) = ( k − 1 ) ( k − 3 ) ⋯ 3 × 1 × E ( X 1 ) = 0 \begin{aligned} E(X^k) &= (k-1)(k-3)\cdots3\times1\times E(X^1) \\ &=0 \end{aligned} E(Xk)=(k1)(k3)3×1×E(X1)=0
综上有:
E ( X k ) = { ∏ i = 1 k / 2 ( 2 i − 1 ) k = 2 i , i = 1 , 2 , 3 ⋯ 0 k = 2 i − 1 E(X^k) = \left \{ \begin{array}{ll} \prod_{i=1}^{k/2}(2i-1) & k=2i, i=1,2,3\cdots \\ & \\ 0 & k=2i-1 \end{array} \right. E(Xk)=i=1k/2(2i1)0k=2i,i=1,2,3k=2i1

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